Question

n
Sum (5r^2+4r-3)
r=1

Answer 1

Step-by-step explanation:

First we factorise the denominator as

2r2+4r+3=2(r+1)(r+3).2r2+4r+3=2(r+1)(r+3).

Next, it makes sense to guess that there exist numbers AA and BB such that we have the identity

2(r+1)(r+3)=Ar+1+Br+3.(1)(1)2(r+1)(r+3)=Ar+1+Br+3.

Forming a common denominator for the right hand side of this, we find that

2(r+1)(r+3)=A(r+3)+B(r+1)(r+1)(r+3)=(A+B)r+3A+B(r+1)(r+3).2(r+1)(r+3)=A(r+3)+B(r+1)(r+1)(r+3)=(A+B)r+3A+B(r+1)(r+3).

Since the numerator must be equal to 22 for all values of rr, we obtain the system of equations

A+B=0,3A+B=2.A+B=0,3A+B=2.

Subtracting these equations gives 2A=22A=2, so A=1A=1 and therefore B=−1B=−1.

Hence

2r2

Answer 2

Answer:

First we factorise the denominator as

2r2+4r+3=2(r+1)(r+3).2r2+4r+3=2(r+1)(r+3).

Next, it makes sense to guess that there exist numbers AA and BB such that we have the identity

2(r+1)(r+3)=Ar+1+Br+3.(1)(1)2(r+1)(r+3)=Ar+1+Br+3.

Forming a common denominator for the right hand side of this, we find that

2(r+1)(r+3)=A(r+3)+B(r+1)(r+1)(r+3)=(A+B)r+3A+B(r+1)(r+3).2(r+1)(r+3)=A(r+3)+B(r+1)(r+1)(r+3)=(A+B)r+3A+B(r+1)(r+3).

Since the numerator must be equal to 22 for all values of rr, we obtain the system of equations

A+B=0,3A+B=2.A+B=0,3A+B=2.

Subtracting these equations gives 2A=22A=2, so A=1A=1 and therefore B=−1B=−1.

Hence

2r2